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How can I prove the that the statements,

(i) $aha^{-1} \in H$ for each $h \in H$

(ii) $a^{-1}ha \in H$ for each $h \in H$

are not equivalent for any particular $a \in G$? I have been trying to come up with a counterexample but I have been unsuccessful. Evidently I would need to find an infinite group which satisfies this, since inductively,

$ \forall h \in H ( aha^{-1} \in H ) \implies \forall h \in H \; \forall n \in \mathbb{N}^+ (a^n h a^{-n} \in H)$

As was pointed out by user Robert Shore.

Kevin Osborn
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  • What @bof meant to type was: is $a^{-1}ha\notin G$ a typo for $a^{-1}ha\notin H$? – Shaun Oct 07 '19 at 23:49
  • I don't believe the statement you're asking a proof for is true in general. If $\ H\ $ is a finite group, for instance, then it's true that for all $\ a\in G\ $ and $\ h\in H\ $ the double implication $$ aha^{-1}\in H\iff a^{-1}ha\in H $$ does hold. – lonza leggiera Oct 07 '19 at 23:52
  • @lonzaleggiera In fact, $ \forall h \in H (aha^{-1} \in H) \Rightarrow \forall h \in H \forall n \in \Bbb N (a^n h a^{-n} \in H) $ so you won't find a countexample where $a$ has finite order. – Robert Shore Oct 08 '19 at 00:09
  • @bof yes that was a typo, my bad! To clarify, here is precisely the wording: Prove the that the statements (i) $ aha^{-1} \in H $ for each $h \in H$ and (ii) $ a^{-1}ha \in H $ for each $h \in H$ are not equivalent for any particular $a \in G$. As I understand it, this would have the quantificational form: $\neg \exists a \in G \forall h \in H ( aha^{-1} \in H \iff a^{-1}ha \in H )$ – Kevin Osborn Oct 08 '19 at 00:29
  • This question has been asked zillions of times before. – Derek Holt Oct 08 '19 at 07:25

1 Answers1

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This question has been asked many times before. A standard counterxample is $$H = \{\left(\begin{array}{cc}1&x\\0&1\end{array}\right) : x \in {\mathbb Z}\} < {\rm GL}(2,{\mathbb Q}),\ \ \ a=\left(\begin{array}{cc}2&0\\0&1\end{array}\right).$$

Derek Holt
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